In the reaction $4Fe + 3O_2 -> 2Fe_2O_3$ , how many grams of $Fe_2O_3$ are produced from $73.2g$ $Fe$ ?

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Answer 1 · 2016-02-27T09:38:30

104.6 g of $"Fe"_2"O"_3$ .

This can be solved through dimensional analysis.

Atomic Masses:

$"Fe = 55.85 g/mol"$
$"O = 16.00 g/mol"$

Formula Mass:

$"Fe"_2"O"_3 ="159.69 g/mol"$

$"Moles of Fe" =73.2cancel("g Fe") xx ("1 mol Fe")/(55.85cancel("g Fe")) = "1.31 mol Fe"$

$"Moles of Fe"_2"O"_3 =1.31 cancel("mol Fe") xx ("2 mol Fe"_2"O"_3)/(4cancel("mol Fe")) = "0.655 mol Fe"_2"O"_3$

$"Mass of Fe"_2"O"_3 = 0.655cancel("mol Fe"_2"O"_3) xx ("159.69 g Fe"_2"O"_3)/(1 cancel("mol Fe"_2"O"_3)) = "104.6 g Fe"_2"O"_3$

In the reaction 4Fe + 3O_2 -> 2Fe_2O_34Fe + 3O_2 -> 2Fe_2O_3, how many grams of Fe_2O_3Fe_2O_3 are produced from 73.2g73.2g FeFe?